Question 1
0/1 pts
africtionless cart is at rest on a tabletop and has a mass of 2.3 kg. it is tied by a
massless string resting on a massless and frictionless pulley to a mass of 1.6 kg
hanging over the edge of the table, (basically the same as what we did in the
lab). calculate the acceleration of the cart assuming that g=9.8 m/s2.

4.0 m/s²

Explanation:

Draw two free body diagrams, one for the cart and one for the hanging mass.

The cart has one force acting on it in the x direction: tension to the right.

The hanging mass has two forces acting on it in the y direction: tension up and weight down.

The hanging mass accelerates downward, so if we take up to be positive, then the acceleration is negative.

Apply Newton's second law to the hanging mass:

∑F = ma

T − m₂g = m₂(-a)

T − m₂g = -m₂a

Apply Newton's second law to the cart:

∑F = ma

T = m₁a

Substitute and solve for a:

m₁a − m₂g = -m₂a

m₁a + m₂a = m₂g

(m₁ + m₂) a = m₂g

a = m₂g / (m₁ + m₂)

Given m₁ = 2.3 kg and m₂ = 1.6 kg:

a = (1.6 kg × 9.8 m/s²) / (2.3 kg + 1.6 kg)

a = 4.0 m/s²

a displacement distance is size of displ

we need to know the time interval during which the force acts. impulse of force = momentum change